Semiconductor physics
Winter term 2012/2013
Prof. Dr. M. S. Brandt
Exercise 13, to be discussed in the week from 28. 1. to 1. 2.
The PNP Transistor as an example of a bipolar transistor
A silicon p-n-p transistor has impurity concentrations of 5 · 1018 cm−3 , 2 · 1017 cm−3 and 1 · 1016
cm−3 in the emitter, base and collector respectively. The base width is 1.0 µm, and the device
cross-sectional area is 0.2 mm2 . The emitter-base junction is forward biased to 0.5 V and the
base-collector junction is reverse biased to 5 V.
a) Calculate the neutral base width.
From the last Problem set we know the built in voltage of a diode to be
Vbi =
kT ND NA
ln
.
q
n2i
(1)
If we solve the poisson equation inside the depletion region and integrate the result once
we arrive at the following fields
qNA (x + WDP )
r 0
qND (WDN − x)
E(x) = −
r 0
E(x) = −
for − WDP ≤ x ≤ 0
(2)
for 0 ≤ x ≤ WDN
(3)
where WDP and WDN are the widths in the n and the p region. If we integrate those
equations again we find the potential across the n and the p region to be
2
qNA WDP
2r 0
2
qND WDN
−Vn =
.
2r 0
Vp =
1
(4)
(5)
If we use NA WDP = Nd WDN and Vbi = Vp − Vn we find
s
2r 0 Vbi ND
WDP =
qNA (NA + ND )
s
2r 0 Vbi NA
WDN =
.
qND (NA + ND )
(6)
(7)
An applied voltage in forward direction reduces the built in voltage and an applied voltage
in the reverse direction adds to the built-in voltage. To calculate the width in the case of
a bias voltage we have to replace Vbi by Vbi − Vbias in the formulas above. All that is left
is determining the values in these formulas. To calculate Vbi we need to calculate
Eg
n2i = NC NV e− kT
(8)
with
1
NC = √
2
1
NV = √
2
m∗e kT
πh̄2
32
m∗h kT
πh̄2
32
(9)
.
(10)
The effective masses of electrons and holes in silicon are m∗e = 1.08m0 and m∗h = 0.81m0
respectively. With this we find
NC = 2.82 · 1019 cm−3
NV = 1.81 · 1019 cm−3
n2i = 1.69 · 1020 cm−6
(11)
(12)
(13)
for a temperature T = 300K. The built in Voltage Vbi is calculated to be 0.9384 V in the
emitter-base junction and 0.795V in the base-collector junction. With this we find the
width of the depletion region of the emitter-base junction in the base to be 5.2 · 10−2 µm
and the width of the base-collector junction in the base to be 4.3 · 10−2 µm. So the total
neutral base width is 1µm − 5.2 · 10−2 µm − 4.3 · 10−2 µm = 0.905µm.
b) Determine the minority carrier concentration at the emitter-base junction.
From the last problem set we know, that the hole concentration at the edge of the depletion layer is
qVEB
kT
pn (0) = pn0 e
n2
pn0 = i .
ND
2
(14)
(15)
If we insert the values for n2i which we have already calculated, we find the hole concentration to be 2.14 · 1011 cm−3 .
2
The diffusion constants of minority carriers in the emitter, the base and the collector are 52 cms ,
2
2
40 cms and 115 cms , respectively and the corresponding lifetimes are 10−8 s, 10−7 s and 10−6 s.
c) How large is the hole current from the emitter to the base and from the base to the
collector?
To calculate the currents we first need the minority carrier distribution in the emitter, base and collector. The minority carrier distribution is found by using the continuity
equation
∂ρ
+ ∇j = 0
∂t
(16)
and adjusting it to the minority carrier densities
Dp
∂ 2 pn pn − pn0
= 0.
−
∂x2
τp
(17)
The solution to this equation is
− Lx
x
pn (x) = pn + C1 e Lp + C2 e
with Lp =
p
(18)
p
Dp τp . In the reverse biased case the boundary conditions are
pn (0) = pn0 e
pn (W ) = 0.
qVEB
kT
(19)
(20)
With these boundary conditions we can find the distribution in the limit of W/Lp 1
to be
x
pn (x) = pn (0)(1 − ).
(21)
W
With the boundary conditions
qVEB
kT
(22)
q|VCB |
− kT
(23)
nE (x = −xe ) = ne0 e
nC (x = xc ) = nc0 e
the same calculation also leads to the distribution of electrons in the emitter
nE (x) = nE0 + nE0 (e
3
qVEB
kT
− 1)e
x+xe
Le
x ≤ −xE
(24)
and the collector
nc (x) = nC0 + nC0 (e−
q|VCB |
kT
c
− x−x
L
− 1)e
C
.
(25)
See Figure 2 for a plot.
In steady state, the current into each area due to minoriy charge carriers has to be equal
to the diffusion of this minority charge right at the boundary to the depletion region. To
Find the hole current into the base, we therefore calculate
qADp pn0 qVEB
∂pn '
e kT = 3.03 · 10−5 A.
(26)
IEp = A −qDp
∂x x=0
W
The neutral base width
p of 0.905µm is much smaller than the diffusion length of the holes
in the base (Lp = Dp τp ≈ 20µm), so we assume that no holes are lost in the base and
set ICp = IEp . This also leads to IBB = 0 since this current is needed to resupply electrons
which recombined with holes in the base.
d) Calculate the electron current from the emitter to the base and from the base to the
collector? Similar to problem c) we calculate the diffusion current at the boundary to the
depletion region.
!
qADE nE0 qVEB
∂nE kT
'
e
− 1 = 1.91 · 10−7 A
(27)
IEn = A −qDe
∂x x=−xE
LE
!
∂nC qADC nC0 −q|VCB |
kT
ICn = A −qDc
'
− 1 = 5.8 · 10−14 A.
(28)
e
∂x x=xC
LC
e) Use your results obtained in problem c) and d) to calculate the total currents into the
emitter, the base and the collector.
IE = IEp + IEn = 3.05 · 10−5 A
IC = ICp + ICn = 3.03 · 10−5 A
IB = IEn + IBB − ICn = 1.9 · 10−7 A
(29)
(30)
(31)
f) Calculate the current gain if the transistor is used in a common emitter configuration.
The current gain is defined by
β=
4
IC
≈ 160
IB
(32)
g) How can the current gain be improved?
– Lower the base width so that even more holes reach the collector.
– Increase the difference in doping between the emitter and the base, so that more
holes are injected in the base.
– Reduce the electron emitter-base-current e.g. by a heterojunction.
A good introduction to bipolar transistors can be found in Sze, ”Semiconductor Devices:
Physics and Technology”.
Figure 1: Currents in a PNP Transistor. Hole currents are blue, electron currents black. From
Sze, ”Semiconductor Devices: Physics and Technology”
Figure 2: Minority charge carrier concentration in the emitter, the base and the collector. From
Sze, ”Semiconductor Devices: Physics and Technology”
5
Figure 3: PNP Transistor in common emitter configuration. From Sze, ”Semiconductor
Devices: Physics and Technology”
Name
e-mail
Tel.Nr.
Office
Prof. Dr. M. S. Brandt MartinS.Brandt@wsi.tum.de 289-12758 S301
Patrick Altmann
patrick.altmann@wsi.tum.de 289-11527 S310
Benedikt Stoib
benedikt.stoib@wsi.tum.de 289-11525 S304
Konrad Klein
konrad.klein@wsi.tum.de
289-11565 S310
Florian Hrubesch
florian.hrubesch@wsi.tum.de 289-11380 C201
The exercises take place on Tue 12:20 - 13:30 (WSI S101), Wed 15:00-17:00 (ZNN 0th floor
seminar room), Thu 08:30-10:00 (WSI S101) and Thu 15:45-17:15 (WSI S101).
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